1 câu trả lời
$$\eqalign{ & \cos 3x - 2\cos 2x = 2 \cr & \Leftrightarrow 4{\cos ^3}x - 3\cos x - 2\left( {2{{\cos }^2}x - 1} \right) = 2 \cr & \Leftrightarrow 4{\cos ^3}x - 3\cos x - 4{\cos ^2}x + 2 = 2 \cr & \Leftrightarrow 4{\cos ^3}x - 4{\cos ^2}x - 3\cos x = 0 \cr & \Leftrightarrow \cos x\left( {4{{\cos }^2}x - 4\cos x - 3} \right) = 0 \cr & \Leftrightarrow \left[ \matrix{ \cos x = 0 \hfill \cr \cos x = {3 \over 2}\,\,\left( {loai} \right) \hfill \cr \cos x = - {1 \over 2} \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = {\pi \over 2} + k\pi \hfill \cr x = \pm {{2\pi } \over 3} + k2\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} $$
